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3 years ago

PLEASE HELP ASAP !!

Chemistry

1 answer:

Ivenika [448]3 years ago

8 0

Answer:

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Theoretical yield of H₂SO₄ = 213 g

percent yield of H₂SO₄ = 94 %

Explanation:

Data Given:

volume of SO₂ = 48.6 L

mass of H₂SO₄ = 200 g

balance equation = ?

theoretical yield = ?

percent yield = ?

Solution:

Part 1:

first we have to write a balance equation for the reaction

SO₂ gas react with water (H₂O) and excess oxygen

The balanced equation is as under

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Part 2:

Now we have to find theoretical yield

First look at the balance reaction

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

2 mol 2 mol

2 moles of SO₂ give gives 2 moles of H₂SO₄

Now calculate volume of 2 moles of SO₂ and mass of 2 moles of H₂SO₄

volume of 2 moles of SO₂

Formula used

volume of gas = no. of moles x molar volume . . . . . . (1)

molar volume of SO₂= 22.4 L/mol

Put values in above formula (1)

volume of gas = 2 mol x 22.4 L/mol

volume of gas = 44.8 L

volume of 2 mole of SO₂ = 44.8 L

Now,

Find mass of 2 mole H₂SO₄

Formula Used

mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of H₂SO₄ = 2 (1) + 32 + 4(16)

molar mass of H₂SO₄ = 98 g/mol

put values in equation 2

mass in grams = 2 mol x 98 g/mol

mass in grams = 196 g

mass of 2 mole of H₂SO₄ = 196 g

** So,

Now we come to know that

44.8 L of SO₂ gives 196 g of H₂SO₄ then how many grams of the H₂SO₄ will be produced by 48.6 L of SO₂

Apply unity Formula

44.8 L of SO₂ ≅ 196 g of H₂SO₄

48.6 L of SO₂ ≅ X g of H₂SO₄

Do cross multiplication

g of H₂SO₄ = 196 g x 48.6 L / 44.8 L

g of H₂SO₄ = 213 g

So that is why the theoretical yield of H₂SO₄ is 213 g

Theoretical yield of H₂SO₄ = 213 g

Part 3

Calculate Percent Yield:

Formula used for this purpose:

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 200 g/ 213 g x 100 %

percent yield = 94 %

So percent yield of H₂SO₄ = 94 %

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b) 1.87

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At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

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The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

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Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

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(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

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Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

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x = 0.0134

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pH = $-log[0.0134]$

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